Question

Find the root of the quadratic equation y=3x^2-30x+45

Answer 1

$y=3x^2-30x+45\\\\a=3;\ b=-30;\ c=45\\\Delta=b^2-4ac\\\\\Delta=(-30)^2-4\cdot3\cdot45=900-540=360 \ \textgreater \ 0\\\\therefore\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{360}=\sqrt{36\cdot10}=\sqrt{36}\cdot\sqrt{10}=6\sqrt{10}\\\\x_1=\dfrac{30-6\sqrt{10}}{2\cdot3}=\dfrac{30-6\sqrt{10}}{6}=\boxed{5-\sqrt{10}}\\\\x_2=\dfrac{30+6\sqrt{10}}{2\cdot3}=\dfrac{30+6\sqrt{10}}{6}=\boxed{5+\sqrt{10}}$